Question: The equation of a circle is given below. $\left(x+\dfrac83 \right)^{2}+y^{2} = 1$ What is its center? $($
Standard equation of the circle A circle is the collection of all points at a distance ${r}$ from a center $({h},{k})$. We can use the Pythagorean theorem to write an equation to relate the center and radius. $({h},{k})$ ${r}$ $x-{h}$ $y-{k}$ $(x, y)$ $\begin{aligned} a^2+b^2&=c^2\\\\ (x - {h})^2 + (y - {k})^2 &= {r}^2 \end{aligned}$ Rewriting the given equation We can rewrite the given equation as: $\begin{aligned}\left(x+\dfrac83 \right)^{2}+y^{2} &= 1\\\\ \left(x - {\left(-\dfrac 83\right)}\right)^2 + (y- 0)^2 &= {1}\end{aligned}$ Finding the center According to the rewritten equation, we can see that the center of the circle is $\left({-\dfrac 83}, {0}\right)$. Finding the radius According to the standard equation of the circle, we get that ${r^2}={1}$. Solving for the radius, we get that $r={\sqrt{1}}={1}$. Summary The circle is centered at $\left(-\dfrac 83, 0\right)$. The circle has a radius of $1$ units.